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S4 Physics Practice Paper: Uniform Acceleration, Newton Laws of Motion and Momentum (100 marks)
Physics – Force and Motion
S4 Physics paper: 6 questions, 18 parts, 100 marks, 60 minutes. Uniform acceleration, Newton second law, momentum, energy conservation, circular motion and percentage error.
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Paper details Level: S4 | Subject: Physics | Time: 60 minutes | Marks: 100 | Questions: 18 | Answers included
Section A – Uniform Acceleration
Answer all three parts. Total 18 marks. A car starts from rest, accelerates uniformly at 2.0 metres per second squared for 8.0 seconds, then brakes at 4.0 metres per second squared until it stops.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | Find the speed after 8.0 seconds and the distance travelled during acceleration. | Answer: Speed = 0 + 2.0 x 8.0 = 16.0 metres per second; distance = 0.5 x 2.0 x 8.0 squared = 64.0 metres. (Marks: speed 3, distance 3) |
| 2 | Find the time taken to stop and the distance travelled while braking. | Answer: Time = 16.0 divided by 4.0 = 4.0 seconds; distance = 16.0 squared divided by (2 x 4.0) = 32.0 metres. (Marks: time 3, distance 3) |
| 3 | Find the average speed for the whole journey and explain why it is not half of the maximum speed. | Answer: Average speed = (64.0 + 32.0) divided by 12.0 = 8.0 metres per second; the accelerating and braking intervals last different times, so the speed change is not uniform over time and half the maximum speed cannot be assumed in general. (Marks: calculation 3, explanation 3) |
Section B – Newton Laws of Motion
Answer all three parts. Total 17 marks. Take the acceleration due to gravity as 10 metres per second squared.
| # | Question | Answer and marking notes |
|---|---|---|
| 4 | A car of mass 1200 kilograms experiences a net force of 3600 newtons. Find its acceleration. | Answer: Acceleration = 3600 divided by 1200 = 3.0 metres per second squared. (Marks: formula 3, value 3) |
| 5 | A car of mass 1400 kilograms is driven by 5000 newtons against a resistance of 800 newtons. Find its acceleration. | Answer: Net force = 5000 – 800 = 4200 newtons; acceleration = 4200 divided by 1400 = 3.0 metres per second squared. (Marks: net force 3, acceleration 3) |
| 6 | A person of mass 60 kilograms stands on a scale in a lift. Find the scale reading when the lift accelerates upwards at 2.0 metres per second squared and when it accelerates downwards at the same rate. | Answer: Upward acceleration gives 60 x (10 + 2.0) = 720 newtons; downward acceleration gives 60 x (10 – 2.0) = 480 newtons. (Marks: upward 3, downward 2) |
Section C – Momentum and Impulse
Answer all three parts. Total 17 marks. A ball of mass 0.15 kilograms strikes a wall horizontally at 20 metres per second and rebounds at 15 metres per second; the contact time is 0.02 seconds.
| # | Question | Answer and marking notes |
|---|---|---|
| 7 | Find the change in momentum of the ball, taking the rebound direction as positive. | Answer: Change in momentum = 0.15 x (20 + 15) = 5.25 kilogram metres per second. (Marks: direction 3, value 3) |
| 8 | Find the average force exerted by the wall on the ball. | Answer: Average force = 5.25 divided by 0.02 = 262.5 newtons. (Marks: formula 3, value 3) |
| 9 | A 2.0 kilogram object moving at 3.0 metres per second collides with a stationary 1.0 kilogram object and they stick together. Find the common speed and the percentage loss of kinetic energy. | Answer: Common speed = (2.0 x 3.0) divided by 3.0 = 2.0 metres per second; kinetic energy falls from 9 joules to 6 joules, a loss of 33.3 percent. (Marks: speed 3, energy loss 2) |
Section D – Work, Energy and Power
Answer all three parts. Total 16 marks. A 2.0 kilogram object slides down a smooth incline of height 5.0 metres; take the acceleration due to gravity as 10 metres per second squared.
| # | Question | Answer and marking notes |
|---|---|---|
| 10 | Find the speed of the object at the bottom of the incline. | Answer: Loss of gravitational potential energy = 2.0 x 10 x 5.0 = 100 joules; speed = square root of (2 x 100 divided by 2.0) = 10.0 metres per second. (Marks: potential energy 3, speed 3) |
| 11 | If friction does 30 joules of work on the incline, find the speed at the bottom. | Answer: Kinetic energy = 100 – 30 = 70 joules; speed = square root of (2 x 70 divided by 2.0) = 8.4 metres per second. (Marks: energy 3, speed 2) |
| 12 | A machine lifts a load at a steady 4.0 metres per second with a constant force of 500 newtons. Find its output power in kilowatts. | Answer: Power = 500 x 4.0 = 2000 watts = 2.0 kilowatts. (Marks: formula 3, unit conversion 2) |
Section E – Circular Motion and Gravity
Answer all three parts. Total 16 marks. Take the Earth radius as 6400 kilometres and surface gravity as 10 metres per second squared.
| # | Question | Answer and marking notes |
|---|---|---|
| 13 | An 800 kilogram car takes a bend of radius 50 metres at 20 metres per second. Find the centripetal force required. | Answer: Centripetal force = 800 x 20 squared divided by 50 = 6400 newtons. (Marks: formula 3, value 3) |
| 14 | Find the acceleration due to gravity at a height of 1600 kilometres and its ratio to the surface value. | Answer: Gravity = 10 x (6400 divided by 8000) squared = 6.4 metres per second squared; ratio = 0.64. (Marks: inverse square 3, ratio 2) |
| 15 | Find the orbital speed of a satellite in a low orbit, using speed squared = gravity times Earth radius. | Answer: Speed = square root of (10 x 6400 x 1000) = 8000 metres per second, that is 8.0 kilometres per second. (Marks: substitution 3, value 2) |
Section F – Experimental Error and Graphs
Answer all three parts. Total 16 marks. In an experiment the acceleration due to gravity is measured with a simple pendulum; the accepted value is 10.0 metres per second squared.
| # | Question | Answer and marking notes |
|---|---|---|
| 16 | The experimental result is 9.8 metres per second squared. Find the percentage error. | Answer: Percentage error = (10.0 – 9.8) divided by 10.0 x 100 percent = 2.0 percent. (Marks: formula 3, value 3) |
| 17 | Three timing readings are 2.4, 2.5 and 2.6 seconds. Find the mean and the range. | Answer: Mean = (2.4 + 2.5 + 2.6) divided by 3 = 2.5 seconds; range = 2.6 – 2.4 = 0.2 seconds. (Marks: mean 3, range 2) |
| 18 | If the reaction time is 0.2 seconds and the trolley moves at 5.0 metres per second, find the distance error caused by timing. Explain how a light gate improves the measurement. | Answer: Distance error = 0.2 x 5.0 = 1.0 metres; a light gate triggers automatically as the trolley passes, so no human reaction is involved and the reaction-time error is removed. (Marks: calculation 3, improvement 2) |
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